Problem: Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x - 19,$ the remainder is 99, and when $P(x)$ is divided by $x - 99,$ the remainder is 19.  What is the remainder when $P(x)$ is divided by $(x - 19)(x - 99)$?
Solution: By the Remainder Theorem, $P(19) = 99$ and $P(99) = 19.$

When $P(x)$ is divided by $(x - 19)(x - 99),$ the remainder must be of the form $ax + b.$  Thus,
\[P(x) = (x - 19)(x - 99) Q(x) + ax + b,\]for some polynomial $Q(x).$

Setting $x = 19$ and $x = 99,$ we get
\begin{align*}
19a + b &= P(19) = 99, \\
99a + b &= P(99) = 19.
\end{align*}Subtracting the equations, we get $80a = -80,$ so $a = -1.$  Then $-19 + b = 99,$ so $b = 118.$  Hence, the remainder is $\boxed{-x + 118}.$